How many ways are there to split a group of 6 boys into two

Question

How many ways are there to split a group of 6 boys into two groups of 3 boys each? (The order of the groups does not matter)

A. 8
B. 10
C. 16
D. 20
E. 24

B 
 Why the answer is not 20? If we select 3 boys out of 6 , we can place them in one group and second group will be automatically selected.

Bunuel · Answer

GENERAL RULE: 1. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is important is \frac{(mn)!}{(n!)^m} 2. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is NOT important is \frac{(mn)!}{(n!)^m*m!} . BACK TO THE ORIGINAL QUESTION: In original question as the order is NOT important, we should use second formula, mn=6 , m=2 groups n=3 objects (people): \frac{(mn)!}{(n!)^m*m!}=\frac{6!}{(3!)^2*2!}=10 . This can be done in another way as well: \frac{C^3_6*C^3_3}{2!}=10 , we are dividing by 2! as there are 2 groups and order doesn't matter. For example if we choose with C^3_6 the group {ABC} then the group {DEF} is left and we have two groups {ABC} and {DEF} but then we could choose also {DEF}, so in this case second group would be {ABC}, so we would have the same two groups: {ABC} and {DEF}. So to get rid of such duplications we should divide C^3_6*C^3_3 by factorial of number of groups - 2!. Answer: B. This concept is discussed at: combinations-problems-95344.html?hilit=dividing%20objects%20order#p734396 Hope it helps.

gurpreetsingh · Answer

thanks I got it

sanjoo · Answer

Bunuel..what if question ask.. order does matter..?? then we wud not divide it by 2! ? can u little elaborate the order does matter and order does not matter?? thank u in advance ..

Bunuel · Answer

Please check these links: 6-people-form-groups-of-2-for-a-practical-work-each-group-95344.html probability-85993.html?highlight=divide+groups combination-55369.html#p690842 probability-88685.html#p669025 combination-groups-and-that-stuff-85707.html#p642634 sub-committee-86346.html?highlight=divide+groups Hope it's clear.

Archit143 · Answer

Hi
 I have a doubt in the question.
 6 students can be arranged into 2 groups of 3 each. so ie 2!
Now these 3 students in each group can be arranged in 3! ways in each group so it gives 3!*3!
So total ways in which the group can be arranged = 2!*3!*3! = 72

Can someone help...where did i go wrong

Archit

KarishmaB · Answer

Responding to a pm: Two groups are not distinct so you don't have 2!. You did not name the groups as GroupA and GroupB. (A, B, C) and (D, E, F) split is the same as (D, E, F) and (A, B, C) split. Also, you do not have to arrange the 3 students in 3! ways. You just have to group them, make a team - not make them stand in a line in a particular sequence. In fact, you can use the opposite method to understand how to get the answer. Arrange all 6 in a line in 6! ways. First 3 is the first group and next 3 is the second group. But guess what, the groups are not distinct so divide by 2!. Also, since the students needn't be arranged, divide by 3! for each group. You get 6!/(2!*3!*3!) = 10 (that's how you get the formula) Also, I have discussed grouping here using different methods: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/1 ... ke-groups/

How many ways are there to split a group of 6 boys into two

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